Biến đổi đơn giản biểu thức chứa căn bậc hai

A \(\begin{array}{l}
a)\,\,6x\sqrt 5 \\
b)\,\,\,\left\{ \begin{array}{l}
\left( {x - y} \right)\sqrt 3 \,\,\,\,khi\,\,\,\,x \ge y\\
\left( {y - x} \right)\sqrt x \,\,\,\,khi\,\,\,x < y
\end{array} \right..
\end{array}\)

B \(\begin{array}{l}
a)\,\,\, - 6x\sqrt 5 \\
b)\,\,\left( {x - y} \right)\sqrt 3
\end{array}\)

C \(\begin{array}{l}
a)\,\,\left\{ \begin{array}{l}
6x\sqrt 5 \,\,\,khi\,\,\,x \ge 0\\
- 6x\sqrt 5 \,\,\,\,khi\,\,\,x < 0
\end{array} \right.\\
b)\,\,\,\left\{ \begin{array}{l}
\left( {x - y} \right)\sqrt 3 \,\,\,khi\,\,\,x \ge 0\\
\left( {y - x} \right)\sqrt 3 \,\,\,khi\,\,\,x < y
\end{array} \right..
\end{array}\)

D \(\begin{array}{l}
a)\,\, - 6x\sqrt 5 \\
b)\,\, - \left( {x - y} \right)\sqrt 3
\end{array}\)

A \(\begin{array}{l}
a)\,\,\left\{ \begin{array}{l}
\frac{{\sqrt 2 }}{2}\,\,\,khi\,\,\,xy > 0\\
- \frac{{\sqrt 2 }}{2}\,\,\,khi\,\,\,xy < 0
\end{array} \right.\\
b)\,\,\left\{ \begin{array}{l}
\sqrt {2{a^2}} \,\,\,\,khi\,\,\,a \ge 0\\
- \sqrt {2{a^2}} \,\,\,\,khi\,\,\,a < 0
\end{array} \right.\\
c)\,\,\sqrt {\frac{a}{b}} \\d)\,\, \sqrt{3} a\\
e)\,\,\left\{ \begin{array}{l}
\sqrt 5 \,\,\,khi\,\,\,x > \frac{1}{2}\\
- \sqrt 5 \,\,\,khi\,\,\,x < \frac{1}{2}
\end{array} \right.
\end{array}\)

B \(\begin{array}{l}
a)\,\,\left\{ \begin{array}{l}
\frac{{\sqrt 2 }}{2}\,\,\,khi\,\,\,xy > 0\\
- \frac{{\sqrt 2 }}{2}\,\,\,khi\,\,\,xy < 0
\end{array} \right.\\
b)\,\,\left\{ \begin{array}{l}
\sqrt {2{a^2}} \,\,\,\,khi\,\,\,a \ge 0\\
- \sqrt {2{a^2}} \,\,\,\,khi\,\,\,a < 0
\end{array} \right.\\
c)\,\, - \sqrt {\frac{a}{b}} \\d)\,\, \sqrt{3a} \\
e)\,\, \sqrt{3} a\\e)\,\,\left\{ \begin{array}{l}\sqrt 5 \,\,\,khi\,\,\,x \ge \frac{1}{2}\\- \sqrt 5 \,\,\,khi\,\,\,x < \frac{1}{2}\end{array} \right.\end{array}\)

C \(\begin{array}{l}
a)\,\,\frac{{\sqrt 2 }}{2}\\
b)\,\,\left\{ \begin{array}{l}
\sqrt {2{a^2}} \,\,\,\,khi\,\,\,a \ge 0\\
- \sqrt {2{a^2}} \,\,\,\,khi\,\,\,a < 0
\end{array} \right.\\
c)\,\,\sqrt {\frac{a}{b}} \\d)\,\, \sqrt{3a} \\
e)\,\,\left\{ \begin{array}{l}
\sqrt 5 \,\,\,khi\,\,\,x \ge \frac{1}{2}\\
- \sqrt 5 \,\,\,khi\,\,\,x < \frac{1}{2}
\end{array} \right.
\end{array}\)

D \(\begin{array}{l}
a)\,\,\frac{{\sqrt 2 }}{2}\\
b)\,\,\left\{ \begin{array}{l}
\sqrt {2{a^2}} \,\,\,\,khi\,\,\,a \ge 0\\
- \sqrt {2{a^2}} \,\,\,\,khi\,\,\,a < 0
\end{array} \right.\\
c)\,\,\sqrt {\frac{a}{b}} \\d)\,\, \sqrt{3} a\\
e)\,\,\sqrt 5
\end{array}\)

A \(\begin{array}{l}
a)\,\,A = - \sqrt {3\left( {y - x} \right)} \\
b)\,\,\,B = 5\sqrt {3x} \\
c)\,\,C = x\sqrt 3 \\
d)\,\,\,D = \sqrt {xy} \left( {10 + 3xy} \right)
\end{array}\)

B \(\begin{array}{l}
a)\,\,A = \sqrt {3\left( {y - x} \right)} \\
b)\,\,\,B = 12\sqrt {3x} \\
c)\,\,C = x\sqrt 3 \\
d)\,\,\,D = \sqrt {xy} \left( {10 + 3xy} \right)
\end{array}\)

C \(\begin{array}{l}
a)\,\,A = - \sqrt {3\left( {y - x} \right)} \\
b)\,\,\,B = 12\sqrt {3x} \\
c)\,\,C = x\sqrt 3 \\
d)\,\,\,D = - \sqrt {xy} \left( {10 + 3xy} \right)
\end{array}\)

D \(\begin{array}{l}
a)\,\,A = - \sqrt {3\left( {y - x} \right)} \\
b)\,\,\,B = 5\sqrt {3x} \\
c)\,\,C = -x\sqrt 3 \\
d)\,\,\,D = \sqrt {xy} \left( {10 + 3xy} \right)
\end{array}\)